Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MX = 4e5 + 69;
const long long inf = 1e18 + 69;
const long long mod = 1e9 + 7;
const long double eps = 1e-7;
int a[222][222], b[222][222];
vector<pair<string, int> > v, pp;
int n, m;
void f(int x, int c) {
  if (c == 0) {
    int mn = 1e9;
    for (int i = 1; i <= m; i++) {
      mn = min(mn, a[x][i]);
    }
    for (int i = 1; i <= m; i++) {
      a[x][i] -= mn;
    }
    for (int i = 0; i < mn; i++) {
      v.push_back({"row", x});
    }
    return;
  } else {
    int mn = 1e9;
    for (int i = 1; i <= n; i++) {
      mn = min(mn, a[i][x]);
    }
    for (int i = 1; i <= n; i++) {
      a[i][x] -= mn;
    }
    for (int i = 0; i < mn; i++) {
      v.push_back({"col", x});
    }
    return;
  }
}
void ff(int x, int c) {
  if (c == 0) {
    int mn = 1e9;
    for (int i = 1; i <= m; i++) {
      mn = min(mn, b[x][i]);
    }
    for (int i = 1; i <= m; i++) {
      b[x][i] -= mn;
    }
    for (int i = 0; i < mn; i++) {
      pp.push_back({"row", x});
    }
    return;
  } else {
    int mn = 1e9;
    for (int i = 1; i <= n; i++) {
      mn = min(mn, b[i][x]);
    }
    for (int i = 1; i <= n; i++) {
      b[i][x] -= mn;
    }
    for (int i = 0; i < mn; i++) {
      pp.push_back({"col", x});
    }
    return;
  }
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      scanf("%d", &a[i][j]);
      b[i][j] = a[i][j];
    }
  }
  for (int i = 1; i <= n; i++) {
    f(i, 0);
  }
  for (int i = 1; i <= m; i++) {
    f(i, 1);
  }
  for (int i = 1; i <= m; i++) {
    ff(i, 1);
  }
  for (int i = 1; i <= n; i++) {
    ff(i, 0);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (a[i][j]) {
        puts("-1");
        return 0;
      }
    }
  }
  if (v.size() < pp.size()) {
    cout << v.size() << endl;
    for (int i = 0; i < v.size(); i++) {
      cout << v[i].first << " " << v[i].second << endl;
    }
  } else {
    cout << pp.size() << endl;
    for (int i = 0; i < pp.size(); i++) {
      cout << pp[i].first << " " << pp[i].second << endl;
    }
  }
  return 0;
}